After T. Piezas answered Is there a series to show $22\pi^4>2143\,$? a natural question is
Is there a series that proves $\frac{22}{7}-\pi>0$?
One such series may be found combining linearly the series that arise from truncating $$\sum_{k=0}^\infty \frac{48}{(4k+3)(4k+5)(4k+7)(4k+9)} = \frac{16}{5}-\pi$$ to two and three terms, namely
$$\sum_{k=2}^\infty \frac{48}{(4 k+3) (4 k+5) (4 k+7) (4 k+9)} = \frac{141616}{45045}-\pi$$ and $$\sum_{k=3}^\infty \frac{48}{(4 k+3) (4 k+5) (4 k+7) (4 k+9)} = \frac{2406464}{765765}-\pi$$ Solving $$a\left(\frac{141616}{45045}-\pi\right)+b\left(\frac{2406464}{765765}-\pi\right)=\frac{22}{7}-\pi$$ for rational $a,b$ and some algebra manipulation yields the result
$$\frac{16}{21} \sum_{k=0}^\infty \frac{1008 k^2+6952 k+12625}{(4 k+11) (4 k+13) (4 k+15) (4 k+17) (4 k+19) (4 k+21)}=\frac{22}{7}-\pi$$
It is interesting to note that the coefficients needed to multiply the two component series are both positive $$a=\frac{113}{7·8·9}$$ $$b=\frac{391}{7·8·9}$$
because the truncation points have been chosen so that
$$\frac{2406464}{765765}<\frac{22}{7}<\frac{141616}{45045}$$
This procedure yields a result that proves the claim with no need for further processing, and it is readily seen to prove $\frac{p}{q}-\pi>0$ for all fractions between $\pi$ and $\frac{16}{5}$.
Now, in the light of this equivalent form of Lehmer's formula $$\pi-3=\sum_{k=1}^\infty \frac{4!}{(4k+1)(4k+2)(4k+4)}$$
one may still ask
Q1 Is there a series that proves $\frac{22}{7}-\pi>0$ with constant numerator?
Q2 Is there a reason why $113$ is both the numerator of the $a$ coefficient and the denominator of the next convergent from above $\frac{355}{113}$?
Edit: A similar series with smaller coefficients may be obtained by applying the method above to $$\begin{align} \sum_{k=0}^\infty \frac{960}{(4 k+3) (4 k+5) (4 k+7) (4 k+9) (4 k+11) (4 k+13)} &= \frac{992}{315}-\pi \\ &= \frac{3·333-7}{3·106-3}-\pi \\ \end{align} $$ in order to obtain $$\sum_{k=0}^\infty \frac{96 (160 k^2+422 k+405)}{(4 k+3) (4 k+5) (4 k+7) (4 k+9) (4 k+11) (4 k+13) (4 k+15) (4 k+17)} = \frac{22}{7}-\pi$$
Q3 What is the relationship between $\frac{992}{315}$ and the third convergent to $\pi$ $\frac{333}{106}$?